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Questions: Algebra BusinessCalculus
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Find the equation of the line perpendicular to \(\displaystyle y = - \frac{x}{10} - 9\) that contains \(\displaystyle \left( -7, \ -9\right)\).
The slope of the line is \(\displaystyle m = - \frac{1}{10}\) so the perpendicular line has slope \(\displaystyle m=10\). The equation has the form \(\displaystyle y=10x+b\). Using the point \(\displaystyle \left( -7, \ -9\right)\) gives the equation \(\displaystyle -9=10\left(-7\right)+b\) Solving for \(\displaystyle b\) gives \(\displaystyle b = 61\). The equation of the perpendicular line is \(\displaystyle y = 10 x + 61\).
\begin{question}Find the equation of the line perpendicular to $y = - \frac{x}{10} - 9$ that contains $\left( -7, \ -9\right)$.
\soln{9cm}{The slope of the line is $m = - \frac{1}{10}$ so the perpendicular line has slope $m=10$. The equation has the form $y=10x+b$. Using the point $\left( -7, \ -9\right)$ gives the equation $-9=10\left(-7\right)+b$ Solving for $b$ gives $b = 61$. The equation of the perpendicular line is $y = 10 x + 61$. }
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Find the equation of the line perpendicular to <img class="equation_image" title=" \displaystyle y = - \frac{x}{10} - 9 " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20-%20%5Cfrac%7Bx%7D%7B10%7D%20-%209%20" alt="LaTeX: \displaystyle y = - \frac{x}{10} - 9 " data-equation-content=" \displaystyle y = - \frac{x}{10} - 9 " /> that contains <img class="equation_image" title=" \displaystyle \left( -7, \ -9\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-7%2C%20%5C%20%20-9%5Cright%29%20" alt="LaTeX: \displaystyle \left( -7, \ -9\right) " data-equation-content=" \displaystyle \left( -7, \ -9\right) " /> . </p> </p><p> <p>The slope of the line is <img class="equation_image" title=" \displaystyle m = - \frac{1}{10} " src="/equation_images/%20%5Cdisplaystyle%20m%20%3D%20-%20%5Cfrac%7B1%7D%7B10%7D%20" alt="LaTeX: \displaystyle m = - \frac{1}{10} " data-equation-content=" \displaystyle m = - \frac{1}{10} " /> so the perpendicular line has slope <img class="equation_image" title=" \displaystyle m=10 " src="/equation_images/%20%5Cdisplaystyle%20m%3D10%20" alt="LaTeX: \displaystyle m=10 " data-equation-content=" \displaystyle m=10 " /> . The equation has the form <img class="equation_image" title=" \displaystyle y=10x+b " src="/equation_images/%20%5Cdisplaystyle%20y%3D10x%2Bb%20" alt="LaTeX: \displaystyle y=10x+b " data-equation-content=" \displaystyle y=10x+b " /> . Using the point <img class="equation_image" title=" \displaystyle \left( -7, \ -9\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cleft%28%20-7%2C%20%5C%20%20-9%5Cright%29%20" alt="LaTeX: \displaystyle \left( -7, \ -9\right) " data-equation-content=" \displaystyle \left( -7, \ -9\right) " /> gives the equation <img class="equation_image" title=" \displaystyle -9=10\left(-7\right)+b " src="/equation_images/%20%5Cdisplaystyle%20-9%3D10%5Cleft%28-7%5Cright%29%2Bb%20" alt="LaTeX: \displaystyle -9=10\left(-7\right)+b " data-equation-content=" \displaystyle -9=10\left(-7\right)+b " /> Solving for <img class="equation_image" title=" \displaystyle b " src="/equation_images/%20%5Cdisplaystyle%20b%20" alt="LaTeX: \displaystyle b " data-equation-content=" \displaystyle b " /> gives <img class="equation_image" title=" \displaystyle b = 61 " src="/equation_images/%20%5Cdisplaystyle%20b%20%3D%2061%20" alt="LaTeX: \displaystyle b = 61 " data-equation-content=" \displaystyle b = 61 " /> . The equation of the perpendicular line is <img class="equation_image" title=" \displaystyle y = 10 x + 61 " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%2010%20x%20%2B%2061%20" alt="LaTeX: \displaystyle y = 10 x + 61 " data-equation-content=" \displaystyle y = 10 x + 61 " /> . </p> </p>